Match List I with List II:

Choose the correct answer from the options given below:

Correct option is D

Given:

List I: Number of trials (n) and probability of success (p)

A. n = 7, p = 0.3

B. n = 5, p = 0.4

C. n = 8, p = 0.5

D. n = 6, p = 0.6

List II: Variance of Binomial Distribution

I. 1.44

II. 2.00

III. 1.20

IV. 1.47

Formula Used:

Variance of a Binomial Distribution = Variance = $$n \cdot p \cdot (1 - p)$$​​

Solution:

Now, the variance for each value in List I:

For A (n = 7, p = 0.3):

Variance = $$7 \cdot 0.3 \cdot (1 - 0.3) = 7 \cdot 0.3 \cdot 0.7 = 1.47$$​​

This matches with IV.

For B (n = 5, p = 0.4):

Variance = $$5 \cdot 0.4 \cdot (1 - 0.4) = 5 \cdot 0.4 \cdot 0.6 = 1.20$$​​

This matches with III.

For C (n = 8, p = 0.5):

Variance = $$8 \cdot 0.5 \cdot (1 - 0.5) = 8 \cdot 0.5 \cdot 0.5 = 2.00$$​​

This matches with II.

For D (n = 6, p = 0.6):

Variance = $$6 \cdot 0.6 \cdot (1 - 0.6) = 6 \cdot 0.6 \cdot 0.4 = 1.44$$​​

This matches with I.

Thus, the correct option is (d) A-IV, B-III, C-II, D-I.